Clearly and as is well known[1]} the \(A^0\)  component of the gauge field \(A^\mu \)  is seen to play the role of a Lagrange multiplier for the first-class constraint
\(\vec{\nabla }\cdot \vec{D}=0\) , namely Gauss' law, while the term \(-\vec{\nabla }\cdot (A^0\vec{D})\)  is a spacelike surface term which at infinity
does not contribute to the total action (when assuming sufficient rapid fall-off of the field \(\vec{D}\) ). However both terms
\(A^0\vec{\nabla }\cdot \vec{D}-\vec{\nabla }\cdot (A^0\vec{D})=-\vec{\nabla }A^0\cdot \vec{D}\)  will be kept in the latter form, for later purposes.
On the other hand the variation of the corresponding action with respect to \(\vec{B}\)  (and accounting for the dependency
of \(\vec{H}\)  on \(\vec{B}\)  such that the Hessian of \({\cal L}_{NLE}(\vec{E},\vec{B})\)  relative to \(\vec{B}\)  be regular) implies once again the equation
\(\vec{B}=\vec{\nabla }\times \vec{A}\) , of which the solution may be used as such by substitution into the above expression, so that the term
in \(\vec{H}\)  no longer contributes to the first-order Lagrangian densityBy considering
from the outset an expression for \({\cal L}^{(1)}_{NLE}\)  which does not include the second term in \(\vec{H}\)  on the r.h.s. of (REF )
one readily reaches the same conclusion for the first-order formulation.. Furthermore the second term in parentheses is an invitation to consider the
following Legendre transformation of \({\cal L}_{NLE}(\vec{E},\vec{B})\)  relative to \(\vec{E}\) , which defines the first-class NLE Hamiltonian density
\({\cal H}_{NLE}\) ,
\({\cal H}_{NLE}(\vec{D},\vec{B})=\vec{E}\cdot \vec{D}-{\cal L}_{NLE}(\vec{E},\vec{B}),\) 
